The perpendicular distance, of the line from the point P(2, –10, 1) is: [2025]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The perpendicular distance, of the line $\frac{x - 1}{2} = \frac{y + 2}{- 1} = \frac{z + 3}{2}$ from the point P(2, –10, 1) is: [2025]

1 6

2 $3 \sqrt{5}$

3 $5 \sqrt{2}$

4 $4 \sqrt{3}$

Ans.
(2)

$\frac{x - 1}{2} = \frac{y + 2}{- 1} = \frac{z + 3}{2} = λ (say)$

Any point on the line is given by

$M (x, y, z) \equiv (2 λ + 1, - λ - 2, 2 λ - 3)$

Also, $\vec{P M} = (2 λ + 1 - 2) \hat{i} + (- λ - 2 + 10) \hat{j} + (2 λ - 3 - 1) \hat{k}$

$= (2 λ - 1) \hat{i} + (- λ + 8) \hat{j} + (2 λ - 4) \hat{k}$ ... (i)

$∵ \vec{P M} \cdot \vec{n} = 0$

$\Rightarrow (2 λ - 1) 2 + (- λ + 8) (- 1) + (2 λ - 4) 2 = 0$

$\Rightarrow 4 λ - 2 + λ - 8 + 4 λ - 8 = 0$

$\Rightarrow 9 λ - 18 = 0 \Rightarrow λ = 2$

$∴ | \vec{P M} | = \sqrt{3^{2} + 6^{2} + 0^{2}} = \sqrt{45} = 3 \sqrt{5}$ [Using (i)]

Want Us to Email you About Special Offers & Updates?